Question

A body drops down a tower and reaches the ground 0.6second . If g=10m/s . Find heigh of the tower and the velocity with which it strikes the ground

Answer 1

Here, u=0 as it is freely falling body
t=0.6sec
From,
$s = ut + \frac{1}{2} at {}^{2}$
Height of the tower =
$\frac{1}{2} \times 10 \times {0.6}^{2} = 18 \: m$
From
$v = u + at$

Final velocity
$v = 10 \times 0.6 = 6 \frac{m}{s}$

Answer 2

Answer:

Explanation: V=0,U=?,g=10,T=0.6 by equation of gravitation. V=U+gT ,putting value in the equation so ,0=U+10×0.6 so U=–6 ,formula of hieght,H=U^2 upon 2g by solving it hieght =1.8m